An organic compound (C_8H_{10}O_2) rotates pl | vedclass

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(A) The molecular formula $C_8H_{10}O_2$ and the positive $FeCl_3$ test indicate that the compound is a phenol derivative.Since the compound rotates p

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$6$

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(A) The molecular formula $C_8H_{10}O_2$ and the positive $FeCl_3$ test indicate that the compound is a phenol derivative.Since the compound rotates plane-polarized light,it must be optically active,meaning it contains a chiral center.The structure consists of a benzene ring with an $-OH$ group and a side chain of $-CH(OH)CH_3$.There are three possible positions for the side chain relative to the $-OH$ group on the benzene ring: ortho,meta,and para.For each of these three positions,the side chain $-CH(OH)CH_3$ contains a chiral carbon atom,which results in two enantiomers ($R$ and $S$ configurations).Therefore,the total number of optically active isomers is $3 	imes 2 = 6$.

An organic compound $(C_8H_{10}O_2)$ rotates plane-polarized light. It produces a pink color with a neutral $FeCl_3$ solution. What is the total number of all the possible isomers for this compound?

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